Ever Wonder why the Mfg. said to use Only their Power Supply ??

[Richard_Cranium72]

You know the little boxes that plug into the wall to power your scanner, printer, modem, battery drill etc ..

If you notice, in the instructions they say something like->>

" CAUTION, USE ONLY XYZ GENUINE REPLACEMENT PARTS ! "

The reason is because they are LIARS !!!!

I checked the power output of several of my little ac/dc power supplies and I was SHOCKED !

The HP for my 5pse scanner said "24 volts DC @ 1.0 amp"
Fine I thought, I check it, and surprise, it is really 22.6 volts.

OK, I say, not too bad.

So I check one marked 13 volts DC
Output->> 15.2 volts DC, **** !!

Still Not TOO BAD.

Checked one on my New Canon Scanner which is marked 10.5 Volts dc.
Output->> 16.2 Volts ,, WHATT ???

Another marked 15.0 vdc
Output->> 19.6 vdc

The only one I checked that was on the money was a Motorola.
Marked 16.0 vdc
Output 15.85 vdc

Test Tool, "Fluke" Digital VOM .

So, If you work on your stuff like most of us do,, CHECK YOUR POWER SUPPLIES,, BEFORE THEY GO OUT..

You may save a headache in replacing it, or at least some funds as the mfg. " GENUINE " replacement parts are usually quite expensive..

DrVette

[Eli]

Have you checked the transformer's outputs when there's a load on them? They're probably well within spec. Try one of those "universal" transformers. Set at 12 volts, I'd bet you'll hit in the 13s somewhere, depending on model, etc.

[jman01pa]

Very true Eli. Even a 12 volt car battery reads much higher until you put a load on it.

J http://sysopt.earthweb.com/forum/smile.gif

[Richard_Cranium72]

Well Gents, Not trying to sound like a smart azz, I've been in auto repair and electrical for over 35 years.

True a auto lead acid battery reads 2.2 vdc per cell = 13.2 vdc approx and during the start cycle drops to sometimes 10.5vdc.

And true also, power supplies can drop when loaded depending upon the circumstances.

HOWEVER, several times, I PERSONALLY have had failures of power supply swapping due to differences in design, output of amperage or voltage swings.

Just a little bitty note, don't take labels at face value, there are often variables that cause things to be different than the indications written.

i.e. The HP power supply,, was LOWER in output than the label indicated, do you perhaps think the voltage will increase with load to the indicated 24 volts? , I'd think not.

ELi has a very valid point that the power supplies do vary in voltage under load,
however I doubt that the one on the Canon Scanner would drop from 16.2 to the label spec of 10.5.

Again, I HAVE encountered problems with power supplies that had specifications that would lend one to think that a substitute should work, when in practice it did not..

DrVette

[Rat]

Another thing to consider: most equipment operates in a wide voltage range, read the equipment docs. A printer with a 12v power supply usually shows a power requirement of 10v - 14v in the specs sheet.

Rat...
"The object of war is not to die for your country but to
make the other ******* die for his."
--George Patton

[tonym]

DrVette,

Eli is on the right track. The "wall wart" power supplies are comprised of a transformer, a rectifier (or bridge) and a filter capacitor. If there is no load, the capacitors peak charge to the maximum level of the secondary's rectified sine wave voltage. Since there is no load, the capacitor won't discharge (much) and the average voltage is essentially the peak of the rectified sine.
When you add a load, the capacitor discharges during the time that the rectified sine wave is below the average value via this added load impedance. In other words, if the R(load) * C time constant is short, then there will be considerable ripple in the time after the sine peak (until the next peak picks up charging the capacitor).

This ripple voltage is worse if the "wall wart" is an 'el-cheapo' and half-wave rectifies vs. full-wave. In the half-wave unit, there is a half-sine peak missing (it is "discarded" by this topology), so there is less capacitor charging per unit time. The result is that the output ripple voltage is greater. The average output voltage is approximately the peak (no load) reading minum half of the ripple voltage value. For a typical wall-wart, this ripple can be on the order of 4 Volts or so, so you can have a 2 Volt decrease in average value from no load to full load!

The deviation is less (*approximately* half of that of a half-wave) for a full-wave unit, but in order to keep costs down, manufacturers choose to save the cost of one or three diodes and suffer with a little higher output ripple.

In the end, any ripple voltage at 60Hz/120Hz is rejected by the voltage regulators in the end-product.

Hope this helps...I wish I could get the hang of pasting an image in my repsonse, so words will have to do!!

[And BTW, if you pry the Motorola unit open, you'll find it has a petite off-line switched-mode power supply crammed in the plastic wart!]


Tony

[This message has been edited by tonym (edited 10-28-2000).]

[Eli]

Remember, the specs are hardly written in stone and there is a range. That's the only explaination for the under-powered transformer.A car battery is a good example, as are other sorts of batteries. Devices will still work when a 1.5 volt alkaline cell drops to 1.4 volts and lower.

[Ted61]

Half wave rectifier? I thought solid state stuff made them obsolete or anti ISO or something like that LOL.

I interchange my power supplies all of the time. That way I don't have to keep track of them.

Ted

[This message has been edited by Ted61 (edited 11-01-2000).]